Question: Is ${39599}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {39599}= &&{3}\cdot10000+ \\&&{9}\cdot1000+ \\&&{5}\cdot100+ \\&&{9}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {39599}= &&{3}(9999+1)+ \\&&{9}(999+1)+ \\&&{5}(99+1)+ \\&&{9}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {39599}= &&\gray{3\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {3}+{9}+{5}+{9}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first four terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${39599}$ is divisible by $3$ if ${ 3}+{9}+{5}+{9}+{9}$ is divisible by $3$ Add the digits of ${39599}$ $ {3}+{9}+{5}+{9}+{9} = {35} $ If ${35}$ is divisible by $3$ , then ${39599}$ must also be divisible by $3$ Add the digits of ${35}$ $ {3}+{5} = \color{#9D38BD}{8} $ If $\color{#9D38BD}{8}$ is divisible by $3$ , then ${35}$ must also be divisible by $3$ $\color{#9D38BD}{8}$ is not divisible by $3$, therefore ${39599}$ must not be divisible by $3$.